Solve 20c^2+37c+15

$$20 c ^ { 2 } + 37 c + 15$$

Factor

$$\left(4c+5\right)\left(5c+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20c^{2}+ac+bc+15. To find a and b, set up a system to be solved.

$$a+b=37 ab=20\times 15=300$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 300.

$$1,300 2,150 3,100 4,75 5,60 6,50 10,30 12,25 15,20$$

Calculate the sum for each pair.

$$1+300=301 2+150=152 3+100=103 4+75=79 5+60=65 6+50=56 10+30=40 12+25=37 15+20=35$$

The solution is the pair that gives sum 37.

$$a=12 b=25$$

Rewrite 20c^{2}+37c+15 as \left(20c^{2}+12c\right)+\left(25c+15\right).

$$\left(20c^{2}+12c\right)+\left(25c+15\right)$$

Factor out 4c in the first and 5 in the second group.

$$4c\left(5c+3\right)+5\left(5c+3\right)$$

Factor out common term 5c+3 by using distributive property.

$$\left(5c+3\right)\left(4c+5\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$20c^{2}+37c+15=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$c=\frac{-37±\sqrt{37^{2}-4\times 20\times 15}}{2\times 20}$$

Square 37.

$$c=\frac{-37±\sqrt{1369-4\times 20\times 15}}{2\times 20}$$

Multiply -4 times 20.

$$c=\frac{-37±\sqrt{1369-80\times 15}}{2\times 20}$$

Multiply -80 times 15.

$$c=\frac{-37±\sqrt{1369-1200}}{2\times 20}$$

Add 1369 to -1200.

$$c=\frac{-37±\sqrt{169}}{2\times 20}$$

Take the square root of 169.

$$c=\frac{-37±13}{2\times 20}$$

Multiply 2 times 20.

$$c=\frac{-37±13}{40}$$

Now solve the equation c=\frac{-37±13}{40} when ± is plus. Add -37 to 13.

$$c=-\frac{24}{40}$$

Reduce the fraction \frac{-24}{40} to lowest terms by extracting and canceling out 8.

$$c=-\frac{3}{5}$$

Now solve the equation c=\frac{-37±13}{40} when ± is minus. Subtract 13 from -37.

$$c=-\frac{50}{40}$$

Reduce the fraction \frac{-50}{40} to lowest terms by extracting and canceling out 10.

$$c=-\frac{5}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{5} for x_{1} and -\frac{5}{4} for x_{2}.

$$20c^{2}+37c+15=20\left(c-\left(-\frac{3}{5}\right)\right)\left(c-\left(-\frac{5}{4}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$20c^{2}+37c+15=20\left(c+\frac{3}{5}\right)\left(c+\frac{5}{4}\right)$$

Add \frac{3}{5} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20c^{2}+37c+15=20\times \frac{5c+3}{5}\left(c+\frac{5}{4}\right)$$

Add \frac{5}{4} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20c^{2}+37c+15=20\times \frac{5c+3}{5}\times \frac{4c+5}{4}$$

Multiply \frac{5c+3}{5} times \frac{4c+5}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20c^{2}+37c+15=20\times \frac{\left(5c+3\right)\left(4c+5\right)}{5\times 4}$$

Multiply 5 times 4.

$$20c^{2}+37c+15=20\times \frac{\left(5c+3\right)\left(4c+5\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20c^{2}+37c+15=\left(5c+3\right)\left(4c+5\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 20

$$x ^ 2 +\frac{37}{20}x +\frac{3}{4} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{37}{20} rs = \frac{3}{4}$$

Two numbers r and s sum up to -\frac{37}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{37}{20} = -\frac{37}{40}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{37}{40} - u s = -\frac{37}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}

$$(-\frac{37}{40} - u) (-\frac{37}{40} + u) = \frac{3}{4}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1369}{1600} - u^2 = \frac{3}{4}$$

Simplify the expression by subtracting \frac{1369}{1600} on both sides

$$-u^2 = \frac{3}{4}-\frac{1369}{1600} = -\frac{169}{1600}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{169}{1600} u = \pm\sqrt{\frac{169}{1600}} = \pm \frac{13}{40}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{37}{40} - \frac{13}{40} = -1.250 s = -\frac{37}{40} + \frac{13}{40} = -0.600$$

Evaluate

$$\left(4c+5\right)\left(5c+3\right)$$